The perpendicular form of the line $\sqrt{3}x-y+4=0$  is

Divide with 2 an both sides we get

$\frac{-\sqrt{3}x}{2}+\frac{1}{2}y=\frac{4}{2}\Rightarrow x\left(\frac{-\sqrt{3}}{2}\right)+y\left(\frac{1}{2}\right)=2$

Here $\cos \alpha =\frac{-\sqrt{3}}{2},\sin \alpha =\frac{1}{2},p=2$

Hence $\alpha$  is in $Q_2\Rightarrow \alpha =\frac{5\pi }{6}$

$\therefore$ The equation n of the line in perpendicular form is $x\cos \frac{5\pi }{6}+y\sin \frac{5\pi }{6}=2$